CHEM 141 1st Edition Lecture 12Outline- Interconverting Hydronium & Hydroxide Concentration- Calculating the pH of a Strong Acid Solution- Calculating the pH of a Strong Base Solution- Diluting a Strong Acid Solution to a Given pH - Preparing a Strong Base Solution with a Given pHInterconverting Hydronium & Hydroxide Concentration- An aqueous solution at 25 degrees Celsius has [OH-]=1.1x10^-7 M, calculate the [H3O+]: Kw=[H3O+][OH-]=1.00x10^-14, plug it in to get [H3O+]=1.00x10^-14/1.1x10^-7, which= 9.1x10^-8M- An aqueous solution at 25 degrees Celsius has [H3O+]=5.3x10^-6, calculate the [OH-]: Kw=[H3O+][OH-]=1.00x10^-14, plug it in to get [OH-]=1.00x10^-14/5.3x10^-6, which=1.9x10^-9M- If a solution has [OH-]=7.69x10^-3M, what is the pH of the solution? Is the solution acidic, basic, or neutral?o pOH=-log[OH-]=-log(7.69x10^-3) this=2.11o pKw=pH+pOH=14o pH=14-2.11=11.89 (basic because over 7)Calculating the pH of a Strong Acid Solution- Calculate the pH of a 0.005M solution of HNO3o HNO3+H2OH3O++NO3-o [H3O+]=[HNO3]=0.005Mo pH=-log[H3O+]=-log(0.005)=2.3Calculating the pH of a Strong Base Solution- Calculate the pH of 0.160M solution of KOHo KOH+H2OK++OH-+H2Oo [KOH]=[OH-]o pOH=-log[OH-]=-log(0.16)=0.8o pKw=pH+pOH=14o pH=14-.8=13.2Diluting a Strong Acid Solution to a Given pH These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- A chemist must prepare 550 mL of hydrobromic acid solution with a pH of 1.2 at 25 degrees Celsius. They will do this in three steps:o 1. Fill a 550mL volumetric flask about halfway with distilled watero 2. Measure out a small volume of concentrated (6.0M) stock solution of HBr and add it to the flasko 3.Fill the flask to the mark with distilled watero Calculate the volume of HBr the chemist must measure out for Step 2 HBr(aq)+H2O(l)H3O+(aq)+Br-(aq) [H3O+]=[HBr] pH=-log[H3O+] [H3O+]=10^-pH=10^-1.2=6.31x10^-2M V1M1=VfMf V1=VfMf/M=0.559(6.31x10^-2)/6M V1=0.00578=5.8mLPreparing a Strong Base Solution with a Given pH- A chemist must prepare 800mL of sodium hydroxide solution with a pH of 12.40 at 25 degrees Celcius. They will do this in three steps:o 1. Fill an 800mL volumetric flask about halfway with distilled watero 2. Measure out a small volume of solid NaOH and add it to the flasko 3. Fill the flask to the mark with distilled watero Calculate the mass of NaOH the chemist must weigh out for step 2 NaOH(s)+H2O(l)Na+(aq)+OH-(aq)+H2O(l) pKw=pH+pOH=14 pOH=14-pH=14-12.4=1.6 pOH=-log[OH-] [OH-]=10^-pOH=10^-1.6=2.51x10^-2 moles=conc x volume (2.51x10^-2)(0.8)=0.02008mol moles=mass/molar mass mass=moles x molar mass of NaOH 0.02008 x
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